Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $x = \dfrac{z^2 + 17z + 72}{9z + 90} \times \dfrac{-5z - 50}{-4z - 32} $
First factor the quadratic. $x = \dfrac{(z + 8)(z + 9)}{9z + 90} \times \dfrac{-5z - 50}{-4z - 32} $ Then factor out any other terms. $x = \dfrac{(z + 8)(z + 9)}{9(z + 10)} \times \dfrac{-5(z + 10)}{-4(z + 8)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (z + 8)(z + 9) \times -5(z + 10) } { 9(z + 10) \times -4(z + 8) } $ $x = \dfrac{ -5(z + 8)(z + 9)(z + 10)}{ -36(z + 10)(z + 8)} $ Notice that $(z + 10)$ and $(z + 8)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ -5\cancel{(z + 8)}(z + 9)(z + 10)}{ -36(z + 10)\cancel{(z + 8)}} $ We are dividing by $z + 8$ , so $z + 8 \neq 0$ Therefore, $z \neq -8$ $x = \dfrac{ -5\cancel{(z + 8)}(z + 9)\cancel{(z + 10)}}{ -36\cancel{(z + 10)}\cancel{(z + 8)}} $ We are dividing by $z + 10$ , so $z + 10 \neq 0$ Therefore, $z \neq -10$ $x = \dfrac{-5(z + 9)}{-36} $ $x = \dfrac{5(z + 9)}{36} ; \space z \neq -8 ; \space z \neq -10 $